[英]Calculate the number of weekdays between 2 dates in R
我正在嘗試編寫一個R函數來計算兩個日期之間的工作日數。 例如,Nweekdays('01 / 30/2011','02/04/2011')等於5。
與此問題類似。 謝謝!
/編輯:@J。 溫徹斯特的答案很棒,但我想知道是否有人能想出一種方法來對此進行矢量化,以便它可以在兩列日期上工作。 謝謝! /編輯2:再次感謝!
Date1 <- as.Date("2011-01-30")
Date2 <- as.Date("2011-02-04")
sum(!weekdays(seq(Date1, Date2, "days")) %in% c("Saturday", "Sunday"))
編輯:和扎克說,讓那里有Vectorize
:)
Dates1 <- as.Date("2011-01-30") + rep(0, 10)
Dates2 <- as.Date("2011-02-04") + seq(0, 9)
Nweekdays <- Vectorize(function(a, b)
sum(!weekdays(seq(a, b, "days")) %in% c("Saturday", "Sunday")))
Nweekdays(Dates1, Dates2)
這些修改后的函數考慮了正數或負數的日期差異,而接受的解決方案考慮了正數日期差異。
library("dplyr")
e2 <- structure(list(date.pr = structure(c(16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16545, 5974), class = "Date"),
date.po = structure(c(16524, 16525, 16526, 16527, 16528, 16529, 16530, 16531, 16538, 16545, 16524, 15974), class = "Date")),
.Names = c("date.1", "date.2"), class = c("tbl_df", "data.frame"), row.names = c(NA, -12L))
1. Locale Dependent Solution: Nweekdays()
函數改編自@J。 贏了。解決方案。 它適用於locale = "English_United States.1252"
Nweekdays <- Vectorize(
function(a, b)
{
ifelse(a < b,
return(sum(!weekdays(seq(a, b, "days")) %in% c("Saturday", "Sunday")) - 1),
return(sum(!weekdays(seq(b, a, "days")) %in% c("Saturday", "Sunday")) - 1))
})
一個。 英語區域
> Sys.setlocale(category="LC_ALL", locale = "English_United States.1252")
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> Sys.getlocale()
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 Monday Monday 0 0
2 2015-03-30 2015-03-31 Monday Tuesday 1 1
3 2015-03-30 2015-04-01 Monday Wednesday 2 2
4 2015-03-30 2015-04-02 Monday Thursday 3 3
5 2015-03-30 2015-04-03 Monday Friday 4 4
6 2015-03-30 2015-04-04 Monday Saturday 5 4
7 2015-03-30 2015-04-05 Monday Sunday 6 4
8 2015-03-30 2015-04-06 Monday Monday 7 5
9 2015-03-30 2015-04-13 Monday Monday 14 10
10 2015-03-30 2015-04-20 Monday Monday 21 15
11 2015-04-20 2015-03-30 Monday Monday 21 15
12 1986-05-11 2013-09-26 Sunday Thursday 10000 7143
灣 中國區域
> Sys.setlocale(category="LC_ALL", locale = "chinese")
[1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936"
> Sys.getlocale()
[1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 0 0
2 2015-03-30 2015-03-31 ÐÇÆÚÒ» ÐÇÆÚ¶þ 1 1
3 2015-03-30 2015-04-01 ÐÇÆÚÒ» ÐÇÆÚÈý 2 2
4 2015-03-30 2015-04-02 ÐÇÆÚÒ» ÐÇÆÚËÄ 3 3
5 2015-03-30 2015-04-03 ÐÇÆÚÒ» ÐÇÆÚÎå 4 4
6 2015-03-30 2015-04-04 ÐÇÆÚÒ» ÐÇÆÚÁù 5 5
7 2015-03-30 2015-04-05 ÐÇÆÚÒ» ÐÇÆÚÈÕ 6 6
8 2015-03-30 2015-04-06 ÐÇÆÚÒ» ÐÇÆÚÒ» 7 7
9 2015-03-30 2015-04-13 ÐÇÆÚÒ» ÐÇÆÚÒ» 14 14
10 2015-03-30 2015-04-20 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 21
11 2015-04-20 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 21
12 1986-05-11 2013-09-26 ÐÇÆÚÈÕ ÐÇÆÚËÄ 10000 10000
2. Locale Independent Solution: Nweekdays()
功能改編自@Sacha Epskamp的解決方案。 它適用於所有語言環境,但@Sacha Epskamp使用c(0,6)
來清除周末,這與使用c(2,3)
提取周末的解決方案不同。
Nweekdays <- Vectorize(
function(a, b) {
return(sum(!(((as.numeric(b:a)) %% 7) %in% c(2,3))) - 1) # 2: Saturday and 3: Sunday
})
一個。 英語區域
> Sys.setlocale(category="LC_ALL", locale = "English_United States.1252")
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> Sys.getlocale()
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 Monday Monday 0 0
2 2015-03-30 2015-03-31 Monday Tuesday 1 1
3 2015-03-30 2015-04-01 Monday Wednesday 2 2
4 2015-03-30 2015-04-02 Monday Thursday 3 3
5 2015-03-30 2015-04-03 Monday Friday 4 4
6 2015-03-30 2015-04-04 Monday Saturday 5 4
7 2015-03-30 2015-04-05 Monday Sunday 6 4
8 2015-03-30 2015-04-06 Monday Monday 7 5
9 2015-03-30 2015-04-13 Monday Monday 14 10
10 2015-03-30 2015-04-20 Monday Monday 21 15
11 2015-04-20 2015-03-30 Monday Monday 21 15
12 1986-05-11 2013-09-26 Sunday Thursday 10000 7143
灣 中國區域
> Sys.setlocale(category="LC_ALL", locale = "chinese")
[1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936"
> Sys.getlocale()
[1] "LC_COLLATE=Chinese (Simplified)_People's Republic of China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic of China.936"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 0 0
2 2015-03-30 2015-03-31 ÐÇÆÚÒ» ÐÇÆÚ¶þ 1 1
3 2015-03-30 2015-04-01 ÐÇÆÚÒ» ÐÇÆÚÈý 2 2
4 2015-03-30 2015-04-02 ÐÇÆÚÒ» ÐÇÆÚËÄ 3 3
5 2015-03-30 2015-04-03 ÐÇÆÚÒ» ÐÇÆÚÎå 4 4
6 2015-03-30 2015-04-04 ÐÇÆÚÒ» ÐÇÆÚÁù 5 4
7 2015-03-30 2015-04-05 ÐÇÆÚÒ» ÐÇÆÚÈÕ 6 4
8 2015-03-30 2015-04-06 ÐÇÆÚÒ» ÐÇÆÚÒ» 7 5
9 2015-03-30 2015-04-13 ÐÇÆÚÒ» ÐÇÆÚÒ» 14 10
10 2015-03-30 2015-04-20 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 15
11 2015-04-20 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 15
12 1986-05-11 2013-09-26 ÐÇÆÚÈÕ ÐÇÆÚËÄ 10000 7143
我寫了這個,但另一個答案更好:)
Nweekdays <- function(a,b)
{
dates <- as.Date(as.Date(a,"%m/%d/%y",origin="1900-01-01"):as.Date(b,"%m/%d/%y",origin="1900-01-01"),origin="1900-01-01")
days <- format(dates,"%w")[c(-1,-length(dates))]
return(sum(!days%in%c(0,6)))
}
Nweekdays('01/30/2011','02/04/2011')
[1] 3
編輯:計算兩個指定日期之間的工作日數。
根據J. Winchesters的建議,該功能可以簡化為:
Nweekdays <- function(a,b)
{
dates <- as.numeric((as.Date(a,"%m/%d/%y")):(as.Date(b,"%m/%d/%y")))
dates <- dates[- c(1,length(dates))]
return(sum(!dates%%7%in%c(0,6)))
}
一些結果:
> Nweekdays('01/30/2011','02/04/2011')
[1] 4
>
> Nweekdays('01/30/2011','01/30/2011')
[1] 0
>
> Nweekdays('01/30/2011','01/25/2011')
[1] 3
請注意,這與區域設置無關。 (關於該主題,我如何更改語言環境?)
我使用以下方法 - 首先是幫手:
weekDays <- function(UPPER = TRUE) {
days <- c('MONDAY', 'TUESDAY', 'WEDNESDAY',
'THURSDAY', 'FRIDAY', 'SATURDAY',
'SUNDAY')
if(!UPPER) return(.Internal(tolower(days)))
days
}
......現在主要功能:
NumWeekDays <- function(dd, Xdays = c('saturday', 'sunday')) {
# a function to count the number of non-Xdays in a month
# >
# first check if Xdays is of correct format
stopifnot( all(.Internal(tolower(Xdays)) %in% weekDays(UP = FALSE)))
# >
# a helper function to find the number of non-X days between two dates
NonXDays <- function(startDate, endDate, Xdays) {
sum(!(.Internal(tolower(weekdays(seq(startDate, endDate, 'day')))) %in%
.Internal(tolower(Xdays))))
}
startDate <- as.Date(as.yearmon(index(dd)), frac = 0)
endDate <- as.Date(as.yearmon(index(dd)), frac = 1)
vapply(1:nrow(dd),
FUN = function(i) NonXDays(startDate[i],
endDate[i],
Xdays = c('saturday', 'sunday')),
FUN.VALUE = numeric(1))
}
例:
set.seed(1)
dx <- apply.monthly(xts(rnorm(600), order.by = Sys.Date() + 1:600), mean)
R> NumWeekDays(dx)
[1] 23 21 22 23 20 23 22 20 22 22 21 22 23 21 22 22 21 23 21 21
使用lubridate解決這個問題,您可以創建如下函數:
library(lubridate)
WorkingDays_function <- function(StartDate,EndDate){
startDate <- dmy(StartDate)
endDate <- dmy(EndDate)
#Now build a sequence between the dates:
myDates <-seq(from = startDate, to = endDate, by = "days")
#Week starts on Sunday (1) so to exclude Sun (1) and Sat (7)
#use > 1 & < 7
working_days <- sum(wday(myDates)>1 & wday(myDates)<7)
print(working_days)
}
WorkingDays_function("11/07/2019","20/07/2019")
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